Question

a manufacturer knows that their items have a normally distributed lifespan, with a mean of 7.2 years, and standard deviation of 1.2 years.
if you randomly purchase one item, what is the probability it will last longer than 9 years?

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 9\), \(\mu=7.2\), and \(\sigma = 1.2\).

$$z=\frac{9 - 7.2}{1.2}=\frac{1.8}{1.2}=1.5$$

Step2: Find the probability

We want \(P(X>9)\), which is equivalent to \(P(Z > 1.5)\) since \(X\sim N(\mu = 7.2,\sigma^{2}=1.2^{2})\) and \(Z=\frac{X-\mu}{\sigma}\sim N(0,1)\).
Using the property \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq1.5)=0.9332\).

$$P(Z > 1.5)=1-0.9332 = 0.0668$$

Answer:

\(0.0668\)